Exam I

Exam I will take place during the lab period on Thursday, Feb. 14 - Happy Valentine's Day. You will have the full three hours, but you shouldn't need that much time. The exam will consist of 6 to 8 problems and cover Chapters 1 through 4.

Below are a few sample problems. You are responsible for all material covered in lecture and homework assignments. DO NOT ASSUME that a subject not covered in the sample exam below will not show up on the exam!

During the exam, you will be provided with an equation sheet. You will be allowed to use only the equations on the sheet, or others that you derive from those on the sheet. You may download the Equation sheet. Resist the temptation to simply look for a similar problem in the text or notes (the standard "get the homework done" technique). That will better prepare you for the exam. If you can't solve the problems, come see me. And don't wait until Wednesday afternoon!!

I've haven't double check my arithmetic, so send me an email if you spot some mistakes.

  1. A person walks east for 100 m, then north for 200 m, and finally 100 m along a diagonal street that is 45o north of east. What is the magnitude and direction of this person's total displacement?

    The x and y components of the displacement is Dx = 100 + 0+ 100 cos45o = 171 m, and Dy = 0 + 200 + 100sin45o = 271 m. Magnitude is sqrt(1712 + 2712) = 320 m at q = tan-1(271/171) = 58o north of east.

  2. You fly from STX directly north to STT at 120 mph in 20 minutes. You then fly directly east to Virgin Gorda at 100 mph for 24 min.
    a) What was your average speed for the entire trip?
    b) What was your average velocity for the entire trip?

    The total distance is 120 mph x 1/3 hr + 100 mph x 2/5 hr = 40 + 40 = 80 mi, traveled in total time 20 + 24 = 44 minutes = 0.73 hr. So ave speed = 80.0 mi/0.73 hr = 109 mph.

    For average velocity, we need the x and y components (Dx, Dy) of the displacement over total time. v = D/Dt = (40, 40) mi /0.73 hr = (55, 55) mph. That is, vx = 55 mph and vy = 55 mph. (Or magnitude 77 mph at 45o north of east.)

  3. Lou Brock stole bases by achieving a full sprint speed of 30 ft/s after only 3 steps (12 feet).
    a) What was his initial acceleration, if we assume it was constant?
    b) What force did the ground exert on Lou? Assume his weight was 170 kg.
    c) How long did it take to steal second base? Assume a total distance of 80 ft and zero acceleration after the first 12 ft.

    Use the convenience equation 302 = 02 + 2a(12 ft) --> a = 37.5 ft/s2. F = ma. So F = (170/32) x 37.5 = 200 lb. The last question must be divided into to parts. During the acc part, 12 = 1/2(37.5)t2 and during the remaining 80-12 = 68 = 30t, so total time = 0.64 + 2.27 = 2.91 sec.

  4. You've landed on the planet Meetzorb. You toss a Meetzorbian rock upward with speed 12 m/s. It returns to your hand 6 seconds later. What is the value of "g" on Meetzorb and how high did the rock go?

    Use symmetry, so it takes 3 seconds to go up and the velocity must -12 m/s when it arrives back at your hand. So -12 = +12 - g(6s) --> g = 4 m/s2. To get the height. y = 12(3)-1/2(4)(32) = 18 m.

  5. Captain Bligh wishes to put a hole at water level in the good ship Lollipop with his cannon. The cannon is 16 ft above the water and fires horizontally. How much gun powder (1 gm for every 1.0 ft/s velocity) does he need if the Lollipop is 300 ft away?

    x = 300 = voxt. voy = 0 so we get t from 0 = 16-1/2(32)t2, so vox = 300/1.0 = 300 ft/s so that's 0.30 kg of powder.

  6. A quarterback throws a pass with speed 50 ft/s at 37o from the horizontal. If the ball is released 6 ft above the ground, how far downfield will it travel before hitting the ground?

    vox = 50 cos37o = 40 ft/s and voy = 50 sin37o = 30 ft/s. 0 = 6 + 30t - 1/2(32)t2 --> t = 2.06 s, so x = 40(2.06) = 82 ft.

  7. Two tugboats are pushing at right angles on a 10,000 kg barge. One is pushing north with force 2.0 x 104N and the other is pushing east with force 4.0 x 104N. What is the magnitude and direction of the acceleration of the barge? Ignore water drag forces.

    F = (4.0 x 104, 2.0 x 104) = 10,000a. So a = (4.0, 2.0) m/s2. Magnitude is sqrt(4.02 + 2.02) = 4.47 m/s2 at q = tan-1(2/4) = 27o north of east.